1991 AHSME Problems/Problem 25

Problem

If $T_n=1+2+3+\cdots +n$ and $P_n=\frac{T_2}{T_2-1}\cdot\frac{T_3}{T_3-1}\cdot\frac{T_4}{T_4-1}\cdot\cdots\cdot\frac{T_n}{T_n-1}$ for $n=2,3,4,\cdots,$ then $P_{1991}$ is closest to which of the following numbers?

$\text{(A) } 2.0\quad \text{(B) } 2.3\quad \text{(C) } 2.6\quad \text{(D) } 2.9\quad \text{(E) } 3.2$

Solution

$\fbox{D}$ It is well known that $T_n = \frac{n(n+1)}{2}$ , so we can put this in and simplify to get that the general term of the product is $\frac{n(n+1)}{(n+2)(n-1)}.$ Now by writing out the terms, we see that almost everything cancels (telescopes), giving $\frac{3 \times 1991}{1993}$ which is almost $3$ , so closest to $2.9.$

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1991 AHSME Problems/Problem 25

Problem

Solution

See also